Question

How do you solve `2x – y = 4`, `3x + y = 1` by graphing and classify the system?

Answer 1

See a solution process below:

Explanation:

For each of the equations we can find two points on the line and then draw a line through the points to graph the line.

Equation 1

For `x = 0`:

`(2 * 0) - y = 4`

`0 - y = 4`

`-y = 4`

`color(red)(-1) xx -y = color(red)(-1) xx 4`

`y = -4` or `(0, -4)`

For `x = 2`

`(2 * 2) - y = 4`

`4 - y = 4`

`-color(red)(4) + 4 - y = -color(red)(4) + 4`

`0 - y = 0`

`-y = 0`

`color(red)(-1) xx -y = color(red)(-1) xx 0`

`y = 0` or `(2, 0)`

graph{(x^2+(y+4)^2-0.05)((x-2)^2+y^2-0.05)(2x-y-4)=0 [-15, 15, -7.5, 7.5]}

Equation 2

For `x = 0`:

`(3 * 0) + y = 1`

`0 + y = 1`

`y = 1` or `(0, 1)`

For `x = 2`

`(3 * 2) + y = 1`

`6 + y = 1`

`-color(red)(6) + 6 + y = -color(red)(6) + 1`

`0 + y = -5`

`y = -5` or `(2, -5)`

graph{(3x+y-1)(x^2+(y-1)^2-0.05)((x-2)^2+(y+5)^2-0.05)(2x-y-4)=0 [-15, 15, -7.5, 7.5]}

Solution

We can see the lines cross at `(1, -2)`

Therefore:

The system is consistent because it has at least one solution. And, it is independent because the lines have different slopes.

graph{(3x+y-1)((x-1)^2+(y+2)^2-0.015)(2x-y-4)=0 [-6, 6, -3, 3]}